Problem: You are given n rectangles, labeled 1 through n. The corners of rectangles have integer coordinates and their edges are parallel to the Ox and Oy axes. The rectangles may touch each other, but they do not overlap (that is, there are no points that belong to the interior of more than one rectangle).

Your task is to determine if there's a non-empty subset of the rectangles that forms a square. That is, determine if there exists a subset of the rectangles and some square for which every point that belongs to the interior or the border of that square belongs to the interior or the border of at least one of the rectangles in the subset, and every point that belongs to the interior or the border of at least one rectangle in the subset belongs to the interior or the border of that square.

Input

First line contains a single integer n (1 ≤ n ≤ 105) — the number of rectangles. Each of the next n lines contains a description of a rectangle, with the i-th such line describing the rectangle labeled i. Each rectangle description consists of four integers: x1, y1, x2, y2 — coordinates of the bottom left and the top right corners (0 ≤ x1 < x2 ≤ 3000, 0 ≤ y1 < y2 ≤ 3000).

No two rectangles overlap (that is, there are no points that belong to the interior of more than one rectangle).

Output

If such a subset exists, print "YES" (without quotes) on the first line of the output file, followed by k, the number of rectangles in the subset. On the second line print k numbers — the labels of rectangles in the subset in any order. If more than one such subset exists, print any one. If no such subset exists, print "NO" (without quotes).

Examples

Input

9
0 0 1 9
1 0 9 1
1 8 9 9
8 1 9 8
2 2 3 6
3 2 7 3
2 6 7 7
5 3 7 6
3 3 5 6


Output

YES 5
5 6 7 8 9


Input

4
0 0 1 9
1 0 9 1
1 8 9 9
8 1 9 8


Output

NO

Note

The first test case looks as follows:

<image>

Note that rectangles 6, 8, and 9 form a square as well, and would be an acceptable answer.

The second test case looks as follows:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxge = 100010;
const int maxtreetop = 8192;
const int maxn = 3001;
struct Trect {
  int x1, y1, x2, y2;
} rect[maxge + 1];
int shang[maxn + 1][maxn + 1], you[maxn + 1][maxn + 1], xia[maxn + 1][maxn + 1],
    zuo[maxn + 1][maxn + 1];
int nextt[maxn + 1], f[maxn + 1];
bool ky[maxn + 1];
int n, ge;
int mianji[maxn + 1][maxn + 1];
int X1, Y1, X2, Y2;
int maxlen = 0;
void init() {
  int i, x1, y1, x2, y2;
  pair<int, pair<int, int> > tmp;
  pair<int, pair<int, pair<int, int> > > qq;
  n = maxn;
  scanf("%d", &ge);
  for (i = 1; i <= ge; ++i) {
    scanf("%d%d%d%d", &rect[i].x1, &rect[i].y1, &rect[i].x2, &rect[i].y2);
    x1 = rect[i].x1;
    y1 = rect[i].y1;
    x2 = rect[i].x2;
    y2 = rect[i].y2;
  }
}
void calc_shang_you() {
  int i, j, k;
  for (i = 1; i <= ge; ++i) {
    for (j = rect[i].y1; j < rect[i].y2; ++j)
      shang[rect[i].x1][j] = shang[rect[i].x2][j] = 1;
    for (j = rect[i].x1; j < rect[i].x2; ++j)
      you[rect[i].y1][j] = you[rect[i].y2][j] = 1;
  }
  for (i = 0; i <= n; ++i) {
    xia[i][1] = shang[i][0];
    for (j = 2; j <= n; ++j) {
      xia[i][j] = shang[i][j - 1];
      if (xia[i][j]) xia[i][j] = 1 + xia[i][j - 1];
    }
    zuo[i][1] = you[i][0];
    for (j = 2; j <= n; ++j) {
      zuo[i][j] = you[i][j - 1];
      if (zuo[i][j]) zuo[i][j] = 1 + zuo[i][j - 1];
    }
  }
  for (i = 0; i <= n; ++i) {
    for (j = n - 1; j >= 0; --j)
      if (shang[i][j]) shang[i][j] = 1 + shang[i][j + 1];
    for (j = n - 1; j >= 0; --j)
      if (you[i][j]) you[i][j] = 1 + you[i][j + 1];
  }
}
void calc_mianji() {
  int i, j, k, sum;
  for (i = 0; i < n; ++i)
    for (j = 0; j < n; ++j) mianji[i][j] = 0;
  for (i = 1; i <= ge; ++i) {
    for (j = rect[i].x1; j < rect[i].x2; ++j)
      for (k = rect[i].y1; k < rect[i].y2; ++k) mianji[j][k] = 1;
  }
  for (i = 0; i < n; ++i) {
    sum = 0;
    for (j = 0; j < n; ++j) {
      sum += mianji[i][j];
      mianji[i][j] = sum + mianji[i - 1][j];
    }
  }
}
void ys(int t) {
  int df, s;
  df = t;
  while (!ky[df]) df = nextt[df];
  while (t != df) {
    s = nextt[t];
    nextt[t] = df;
    t = s;
  }
}
int calcmj(int x1, int y2, int x2, int y1) {
  int q, q2 = 0, q3 = 0, q4 = 0;
  q = mianji[x2 - 1][y2 - 1];
  if (x1 >= 0) q3 = mianji[x1 - 1][y2 - 1];
  if (y1 >= 0) q2 = mianji[x2 - 1][y1 - 1];
  if ((x1 >= 0) && (y1 >= 0)) q4 = mianji[x1 - 1][y1 - 1];
  return q - q2 - q3 + q4;
}
void calc_result() {
  int x, y, t, start, stop, l, df, s;
  for (int w = 1; w <= n + n - 1; ++w) {
    if (w <= n) {
      x = 0;
      y = w;
    } else {
      x = w - n;
      y = n;
    }
    start = x;
    while ((x <= n) && (y >= 0)) {
      f[x] = xia[x][y];
      if (you[y][x] < f[x]) f[x] = you[y][x];
      ky[x] = true;
      if (x > start) {
        nextt[x - 1] = x;
        l = zuo[y][x];
        if (shang[x][y] < l) l = shang[x][y];
        if (l > 0) {
          t = x - l;
          if (t < start) t = start;
          if (!ky[t]) {
            ys(t);
            t = nextt[t];
          }
          while (t < x) {
            if (!ky[t]) {
              ys(t);
              t = nextt[t];
            }
            if (t < x) {
              if (t + f[t] >= x) {
                if (calcmj(t, x + y - t, x, y) < (x - t) * (x - t)) {
                  ky[t] = false;
                  t = nextt[t];
                  start = t + 1;
                } else {
                  if (x - t > maxlen) {
                    maxlen = x - t;
                    X1 = t;
                    Y1 = y;
                    X2 = x;
                    Y2 = x + y - t;
                  }
                  break;
                }
              } else {
                ky[t] = false;
                t = nextt[t];
              }
            } else
              break;
          }
        }
      }
      ++x;
      --y;
    }
  }
}
void work() {
  calc_shang_you();
  calc_mianji();
  calc_result();
}
void output() {
  int i, cnt;
  bool first = true;
  if (maxlen == 0)
    printf("NO\n");
  else {
    printf("YES ");
    cnt = 0;
    for (i = 1; i <= ge; ++i)
      if ((rect[i].x1 >= X1) && (rect[i].x2 <= X2) && (rect[i].y1 >= Y1) &&
          (rect[i].y2 <= Y2))
        ++cnt;
    printf("%d\n", cnt);
    for (i = 1; i <= ge; ++i)
      if ((rect[i].x1 >= X1) && (rect[i].x2 <= X2) && (rect[i].y1 >= Y1) &&
          (rect[i].y2 <= Y2)) {
        if (first)
          first = false;
        else
          printf(" ");
        printf("%d", i);
      }
    printf("\n");
  }
}
int main() {
  init();
  work();
  output();
  return 0;
}